Problem: Simplify and expand the following expression: $ \dfrac{3}{t + 4}- \dfrac{1}{4t + 36}+ \dfrac{4}{t^2 + 13t + 36} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4t + 36} = \dfrac{1}{4(t + 9)}$ We can factor the quadratic in the third term: $ \dfrac{4}{t^2 + 13t + 36} = \dfrac{4}{(t + 4)(t + 9)}$ Now we have: $ \dfrac{3}{t + 4}- \dfrac{1}{4(t + 9)}+ \dfrac{4}{(t + 4)(t + 9)} $ The least common multiple of the denominators is: $ (t + 4)(t + 9)$ In order to get the first term over $(t + 4)(t + 9)$ , multiply by $\dfrac{4(t + 9)}{4(t + 9)}$ $ \dfrac{3}{t + 4} \times \dfrac{4(t + 9)}{4(t + 9)} = \dfrac{12(t + 9)}{(t + 4)(t + 9)} $ In order to get the second term over $(t + 4)(t + 9)$ , multiply by $\dfrac{t + 4}{t + 4}$ $ \dfrac{1}{4(t + 9)} \times \dfrac{t + 4}{t + 4} = \dfrac{t + 4}{(t + 4)(t + 9)} $ In order to get the third term over $(t + 4)(t + 9)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4}{(t + 4)(t + 9)} \times \dfrac{4}{4} = \dfrac{16}{(t + 4)(t + 9)} $ Now we have: $ \dfrac{12(t + 9)}{(t + 4)(t + 9)} - \dfrac{t + 4}{(t + 4)(t + 9)} + \dfrac{16}{(t + 4)(t + 9)} $ $ = \dfrac{ 12(t + 9) - (t + 4) + 16} {(t + 4)(t + 9)} $ Expand: $ = \dfrac{12t + 108 - t - 4 + 16}{4t^2 + 52t + 144} $ $ = \dfrac{11t + 120}{4t^2 + 52t + 144}$